3.49.26 \(\int \frac {-2+6 x-2 x^2+(3 x^2-x^3+(3-x) \log (\frac {1}{4} (9-6 x+x^2))) \log (x^2+\log (\frac {1}{4} (9-6 x+x^2)))}{(-3 x^2+x^3+(-3+x) \log (\frac {1}{4} (9-6 x+x^2))) \log (x^2+\log (\frac {1}{4} (9-6 x+x^2)))} \, dx\)

Optimal. Leaf size=23 \[ 6-x-\log \left (\log \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right )\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 2, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6688, 6684} \begin {gather*} -\log \left (\log \left (x^2+\log \left (\frac {1}{4} (3-x)^2\right )\right )\right )-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + 6*x - 2*x^2 + (3*x^2 - x^3 + (3 - x)*Log[(9 - 6*x + x^2)/4])*Log[x^2 + Log[(9 - 6*x + x^2)/4]])/((-3
*x^2 + x^3 + (-3 + x)*Log[(9 - 6*x + x^2)/4])*Log[x^2 + Log[(9 - 6*x + x^2)/4]]),x]

[Out]

-x - Log[Log[x^2 + Log[(3 - x)^2/4]]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1-\frac {2 \left (1-3 x+x^2\right )}{(-3+x) \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right )}\right ) \, dx\\ &=-x-2 \int \frac {1-3 x+x^2}{(-3+x) \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right ) \log \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right )} \, dx\\ &=-x-\log \left (\log \left (x^2+\log \left (\frac {1}{4} (3-x)^2\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 22, normalized size = 0.96 \begin {gather*} -x-\log \left (\log \left (x^2+\log \left (\frac {1}{4} (-3+x)^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 6*x - 2*x^2 + (3*x^2 - x^3 + (3 - x)*Log[(9 - 6*x + x^2)/4])*Log[x^2 + Log[(9 - 6*x + x^2)/4]]
)/((-3*x^2 + x^3 + (-3 + x)*Log[(9 - 6*x + x^2)/4])*Log[x^2 + Log[(9 - 6*x + x^2)/4]]),x]

[Out]

-x - Log[Log[x^2 + Log[(-3 + x)^2/4]]]

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fricas [A]  time = 1.08, size = 23, normalized size = 1.00 \begin {gather*} -x - \log \left (\log \left (x^{2} + \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*log(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*log(log(1/4*x^2-3/2*x+9/4)+x^2)-2*x^2+6*x-2)/((x-3)*log(1/
4*x^2-3/2*x+9/4)+x^3-3*x^2)/log(log(1/4*x^2-3/2*x+9/4)+x^2),x, algorithm="fricas")

[Out]

-x - log(log(x^2 + log(1/4*x^2 - 3/2*x + 9/4)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, x^{2} + {\left (x^{3} - 3 \, x^{2} + {\left (x - 3\right )} \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )} \log \left (x^{2} + \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right ) - 6 \, x + 2}{{\left (x^{3} - 3 \, x^{2} + {\left (x - 3\right )} \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )} \log \left (x^{2} + \log \left (\frac {1}{4} \, x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right )\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*log(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*log(log(1/4*x^2-3/2*x+9/4)+x^2)-2*x^2+6*x-2)/((x-3)*log(1/
4*x^2-3/2*x+9/4)+x^3-3*x^2)/log(log(1/4*x^2-3/2*x+9/4)+x^2),x, algorithm="giac")

[Out]

integrate(-(2*x^2 + (x^3 - 3*x^2 + (x - 3)*log(1/4*x^2 - 3/2*x + 9/4))*log(x^2 + log(1/4*x^2 - 3/2*x + 9/4)) -
6*x + 2)/((x^3 - 3*x^2 + (x - 3)*log(1/4*x^2 - 3/2*x + 9/4))*log(x^2 + log(1/4*x^2 - 3/2*x + 9/4))), x)

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maple [A]  time = 0.06, size = 24, normalized size = 1.04

method	result	size
norman	\(-x -\ln \left (\ln \left (\ln \left (\frac {1}{4} x^{2}-\frac {3}{2} x +\frac {9}{4}\right )+x^{2}\right )\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3-x)*ln(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*ln(ln(1/4*x^2-3/2*x+9/4)+x^2)-2*x^2+6*x-2)/((x-3)*ln(1/4*x^2-3/2*
x+9/4)+x^3-3*x^2)/ln(ln(1/4*x^2-3/2*x+9/4)+x^2),x,method=_RETURNVERBOSE)

[Out]

-x-ln(ln(ln(1/4*x^2-3/2*x+9/4)+x^2))

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maxima [A]  time = 0.49, size = 22, normalized size = 0.96 \begin {gather*} -x - \log \left (\log \left (x^{2} - 2 \, \log \relax (2) + 2 \, \log \left (x - 3\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*log(1/4*x^2-3/2*x+9/4)-x^3+3*x^2)*log(log(1/4*x^2-3/2*x+9/4)+x^2)-2*x^2+6*x-2)/((x-3)*log(1/
4*x^2-3/2*x+9/4)+x^3-3*x^2)/log(log(1/4*x^2-3/2*x+9/4)+x^2),x, algorithm="maxima")

[Out]

-x - log(log(x^2 - 2*log(2) + 2*log(x - 3)))

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mupad [B]  time = 3.80, size = 23, normalized size = 1.00 \begin {gather*} -x-\ln \left (\ln \left (\ln \left (\frac {x^2}{4}-\frac {3\,x}{2}+\frac {9}{4}\right )+x^2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(x^2/4 - (3*x)/2 + 9/4) + x^2)*(log(x^2/4 - (3*x)/2 + 9/4)*(x - 3) - 3*x^2 + x^3) - 6*x + 2*x^2 +
2)/(log(log(x^2/4 - (3*x)/2 + 9/4) + x^2)*(log(x^2/4 - (3*x)/2 + 9/4)*(x - 3) - 3*x^2 + x^3)),x)

[Out]

- x - log(log(log(x^2/4 - (3*x)/2 + 9/4) + x^2))

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sympy [A]  time = 0.50, size = 24, normalized size = 1.04 \begin {gather*} - x - \log {\left (\log {\left (x^{2} + \log {\left (\frac {x^{2}}{4} - \frac {3 x}{2} + \frac {9}{4} \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*ln(1/4*x**2-3/2*x+9/4)-x**3+3*x**2)*ln(ln(1/4*x**2-3/2*x+9/4)+x**2)-2*x**2+6*x-2)/((x-3)*ln(
1/4*x**2-3/2*x+9/4)+x**3-3*x**2)/ln(ln(1/4*x**2-3/2*x+9/4)+x**2),x)

[Out]

-x - log(log(x**2 + log(x**2/4 - 3*x/2 + 9/4)))

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